0
\$\begingroup\$

Background: In doing some experimenting with low-voltage home lighting, I have an application that requires a resistor (or else LED lights become damaged) and I have also, through trial-and-error, discovered the correct resistance to use in this application.

Everything works - except - the single resistor I am using is overheating.

I know how to use a multimeter, so I can measure the resistances involved, but I bought these resistors in bulk, so I have no idea of the specific stats of what power level they are rated for, etc.

I have some very basic-level understanding of electronics, and a theory on how to avoid this problem so that I can use my setup safely.

Question: Would it be a viable strategy to replace my single resistor with multiple resistors connected in series, where the sum of the resistances add up to the value of the original resistor?

If I am understanding this correctly, even though the current (which in my head I imagine as number of electrons per second) flowing would be the same, the fact that there is a lower "voltage drop" across each individual resistor would reduce the power dissipation across each resistor, and result in less heat being produced?

Thanks for your answers and I will be happy make edits to my question to put it in the form where it can be more helpful to others in the future.

\$\endgroup\$

marked as duplicate by Chris Stratton, Sparky256, PeterJ, laptop2d, winny Jan 9 '18 at 11:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 3
    \$\begingroup\$ Resistors are not the proper solution for LED lighting, they should only be used for low power indicator LEDs. For LED lighting, you need to use efficient current-mode drivers, otherwise the power wasted in the resistors cuts into the energy savings you hoped to gain by using LEDs in the first place. \$\endgroup\$ – Chris Stratton Dec 29 '17 at 2:32
  • \$\begingroup\$ Yes. Let's consider a made-up example. If you put two resistors in series, each one rated at 1 Ohm and 0.5W, that is the same as having one resistor rated at 2 Ohm and 1W. Similarly, if you put two resistors in parallel, each one rated at 4 Ohm and 0.5W, that is the same as having one resistor rated at 2 Ohm and 1W. Things can get more complicated if the resistors are not the same value, or if you mix them in more complicated ways. \$\endgroup\$ – mkeith Dec 29 '17 at 2:37
  • \$\begingroup\$ It would help a lot if you could post a picture, and/or a schematic, and/or let us know the value of the resistor in question, and how much voltage is across it before it burns out. \$\endgroup\$ – mkeith Dec 29 '17 at 2:38
  • \$\begingroup\$ Charges per second is a good way to visualize current. If the "charge" happens to be measured in the unit of "Coulombs" then you get Coulombs/sec which is the definition of the Ampere (or Amp). It takes a whole bunch of electrons to add up to a Coulomb. \$\endgroup\$ – mkeith Dec 29 '17 at 2:44
  • 1
    \$\begingroup\$ If you start dumping more power into resistors than the LED's then your supply voltage does not match the load . It is better to define the Lights and power source and ask for the formula to compute the optimum value and rating. \$\endgroup\$ – Sunnyskyguy EE75 Dec 29 '17 at 3:46
3
\$\begingroup\$

You are talking about splitting the power dissipated by the resistor across multiple resistors. This is a fairly normal and valid approach.

You actually have multiple choices of ways to do that, three of which are shown below.

  1. Two resistors, half the value connected in series.

  2. Two resistors double the value connected in parallel.

  3. Two strings of two resistors of the same value in series, connected in parallel. (Splits the power over 4)

schematic

simulate this circuit – Schematic created using CircuitLab

However, I am concerned this is really an X-Y problem, and your resistor approach may not be the best one. Especially if you are dissipating that much heat through them.

\$\endgroup\$
  • 1
    \$\begingroup\$ of the three options, the best for a quick kludge is #3, because it uses the original value resistors. \$\endgroup\$ – Neil_UK Dec 29 '17 at 7:43
  • 1
    \$\begingroup\$ This is a fantastic answer and I appreciate the level of detail. Since I am dealing with resistors purchased in bulk, I do in fact have a bunch with the same value "R" already selected and ready for use. I'm sure I have all the others too, but it means delving into the unknown piles of resistors with the multi-meter :) So option 3 is what I will be using. Sorry I lost track of this question due to the marking as duplicate. \$\endgroup\$ – user1445967 Mar 13 '18 at 5:31
1
\$\begingroup\$

Power (or heat produced) = Voltage times current (P=VI), and Ohm's Law says Voltage = current x resistance (V=IR) (where I = current, in Amperes).

So, if you connect several resistors in series of the same value you are currently using, you will both reduce the voltage across any one resistor, and reduce the current through all resistors and your LEDs. If you wish to maintain the current (and so the same brightness of the LED), you could use four resistors of 1/4 the value of your present resistor - this would divide the power dissipation over four resistors, making each one less hot.

Alternatively, you could connect several resistors in parallel, perhaps four resistors each of four times the value of your present resistor (to keep the same effective resistance) - this would also split the power dissipation over four resistors.

(I only picked four as a handy example number - you could use any number of resistors...)

\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.